3.158 \(\int \frac {x^4}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=99 \[ \frac {256 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac {32 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {16 x^3}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

-2/3*x^4/b/arctanh(tanh(b*x+a))^(3/2)-128/3*x*arctanh(tanh(b*x+a))^(3/2)/b^4+256/15*arctanh(tanh(b*x+a))^(5/2)
/b^5-16/3*x^3/b^2/arctanh(tanh(b*x+a))^(1/2)+32*x^2*arctanh(tanh(b*x+a))^(1/2)/b^3

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Rubi [A]  time = 0.07, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2157, 30} \[ -\frac {16 x^3}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {32 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^5}-\frac {2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(-2*x^4)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (16*x^3)/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (32*x^2*Sqrt[Arc
Tanh[Tanh[a + b*x]]])/b^3 - (128*x*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b^4) + (256*ArcTanh[Tanh[a + b*x]]^(5/2))/
(15*b^5)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {x^4}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac {2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {8 \int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 b}\\ &=-\frac {2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {16 x^3}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 \int \frac {x^2}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b^2}\\ &=-\frac {2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {16 x^3}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {32 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {64 \int x \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=-\frac {2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {16 x^3}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {32 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac {128 \int \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b^4}\\ &=-\frac {2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {16 x^3}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {32 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac {128 \operatorname {Subst}\left (\int x^{3/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^5}\\ &=-\frac {2 x^4}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {16 x^3}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {32 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 83, normalized size = 0.84 \[ -\frac {2 \left (40 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))-240 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+320 b x \tanh ^{-1}(\tanh (a+b x))^3-128 \tanh ^{-1}(\tanh (a+b x))^4+5 b^4 x^4\right )}{15 b^5 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(-2*(5*b^4*x^4 + 40*b^3*x^3*ArcTanh[Tanh[a + b*x]] - 240*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 320*b*x*ArcTanh[Ta
nh[a + b*x]]^3 - 128*ArcTanh[Tanh[a + b*x]]^4))/(15*b^5*ArcTanh[Tanh[a + b*x]]^(3/2))

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fricas [A]  time = 0.49, size = 74, normalized size = 0.75 \[ \frac {2 \, {\left (3 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} + 192 \, a^{3} b x + 128 \, a^{4}\right )} \sqrt {b x + a}}{15 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*b^4*x^4 - 8*a*b^3*x^3 + 48*a^2*b^2*x^2 + 192*a^3*b*x + 128*a^4)*sqrt(b*x + a)/(b^7*x^2 + 2*a*b^6*x + a
^2*b^5)

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giac [A]  time = 0.25, size = 75, normalized size = 0.76 \[ \frac {2 \, {\left (12 \, {\left (b x + a\right )} a^{3} - a^{4}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{5}} + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{20} - 20 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{20} + 90 \, \sqrt {b x + a} a^{2} b^{20}\right )}}{15 \, b^{25}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

2/3*(12*(b*x + a)*a^3 - a^4)/((b*x + a)^(3/2)*b^5) + 2/15*(3*(b*x + a)^(5/2)*b^20 - 20*(b*x + a)^(3/2)*a*b^20
+ 90*sqrt(b*x + a)*a^2*b^20)/b^25

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maple [B]  time = 0.15, size = 295, normalized size = 2.98 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}-\frac {8 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} a}{3}-\frac {8 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3}+12 a^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}+24 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}+12 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (-4 a^{3}-12 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )-12 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {2 \left (a^{4}+4 a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+6 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+4 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}\right )}{3 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

2/b^5*(1/5*arctanh(tanh(b*x+a))^(5/2)-4/3*arctanh(tanh(b*x+a))^(3/2)*a-4/3*arctanh(tanh(b*x+a))^(3/2)*(arctanh
(tanh(b*x+a))-b*x-a)+6*a^2*arctanh(tanh(b*x+a))^(1/2)+12*a*(arctanh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(
1/2)+6*(arctanh(tanh(b*x+a))-b*x-a)^2*arctanh(tanh(b*x+a))^(1/2)-(-4*a^3-12*a^2*(arctanh(tanh(b*x+a))-b*x-a)-1
2*a*(arctanh(tanh(b*x+a))-b*x-a)^2-4*(arctanh(tanh(b*x+a))-b*x-a)^3)/arctanh(tanh(b*x+a))^(1/2)-1/3*(a^4+4*a^3
*(arctanh(tanh(b*x+a))-b*x-a)+6*a^2*(arctanh(tanh(b*x+a))-b*x-a)^2+4*a*(arctanh(tanh(b*x+a))-b*x-a)^3+(arctanh
(tanh(b*x+a))-b*x-a)^4)/arctanh(tanh(b*x+a))^(3/2))

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maxima [A]  time = 0.52, size = 64, normalized size = 0.65 \[ \frac {2 \, {\left (3 \, b^{5} x^{5} - 5 \, a b^{4} x^{4} + 40 \, a^{2} b^{3} x^{3} + 240 \, a^{3} b^{2} x^{2} + 320 \, a^{4} b x + 128 \, a^{5}\right )}}{15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/15*(3*b^5*x^5 - 5*a*b^4*x^4 + 40*a^2*b^3*x^3 + 240*a^3*b^2*x^2 + 320*a^4*b*x + 128*a^5)/((b*x + a)^(5/2)*b^5
)

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mupad [B]  time = 1.42, size = 817, normalized size = 8.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/atanh(tanh(a + b*x))^(5/2),x)

[Out]

((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((3*(lo
g(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/(2*b^4) +
(2*((2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/b^
3 + (8*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/
(5*b^3))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)
)/(3*b)))/b + (2*x^2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) +
1))/2)^(1/2))/(5*b^3) + (x*((2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2
*b*x) + 1)) + 2*b*x))/b^3 + (8*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp
(2*b*x) + 1))/2 + b*x))/(5*b^3))*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*e
xp(2*b*x) + 1))/2)^(1/2))/(3*b) - (2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*
a)*exp(2*b*x) + 1))/2)^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b
*x) + 1)) + 2*b*x)^3)/(b^5*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x
) + 1)))) - ((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(
1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4)/(6
*b^5*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Integral(x**4/atanh(tanh(a + b*x))**(5/2), x)

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